## aroub 3 years ago Why do we have to put an absolute value in the final answer?!

1. estudier

2. aroub

$\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}$

3. estudier

Can't even see that, is it 10 th root?

4. aroub

Yup

5. akash_809

$\sqrt[10]{x^10}$

6. akash_809

see moderator has warned me now...whats the way to delet ur account...can somebody give the link...i m going frm here

7. aroub

What? O.o

8. akash_809

can u pls give me the link...i want to delet my account..i can't find it anywhere

9. aroub

I don't know.. You have to contact a moderator

10. akash_809

cmon moderator show up now...

11. phi

The only reason I can think of is they want to make sure you get a real and postiive (as opposed to complex) root. There will be 10 different roots.

12. TuringTest

@akash_809 you cannot delete OS accounts

13. aroub

But you can write it without absolute value, right? Because (or this is what I've been taught) there's x^3 in the radicand so it's definitely positive..

14. TuringTest

the simple reason is that$x^{10}=(-x)^{10}$so $\large\sqrt[10]{x^{10}}$will give you the same answer whether x is negative or positive

15. akash_809

what...thats cynical...open study community is surely going nowher with all these warnings and suspensions...and to top that off , you even cant delete your account...what is this open study team please response...i wnat to know how can i delete my account

16. TuringTest

absolute value of x is actually defined in 2 ways: 1) distance from the point x=0 on the number line 2) $$|x|=\sqrt{x^2}$$

17. aroub

Thanks :))

18. TuringTest

so since$\sqrt{(-x)^2}=\sqrt{x^2}=|x|$then$\sqrt[10]{x^{10}}=\sqrt[10]{(-x)^{10}}=|x|$(because x to any even power is positive, whether x>0 or x<0)

19. TuringTest

20. TuringTest

welcome :D

21. aroub

In this case I totally understand why. But in the question itself there's an odd exponent in the radicand so it's definitely positive right? So then no need for an absolute value..

22. akash_809

@TuringTest root of negative number is not defined...it will result in iota value...so -x wont be the answer i think

23. TuringTest

not necessarily, let's assume x<0...

24. aroub

If you assumed that, the question will be wrong.

25. aroub

-I think-

26. akash_809

yes it is...$lets take an example......\sqrt{-2} is \not defined unless you are taking imaginary numbers into account then u wl get +-1.414 i$

27. TuringTest

the root of a negative number is not undefined, it is imaginary...$\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}$

28. akash_809

sry reply got messed up i was just saying that square root of -2 is not defined unless you consider imaginary numbers as wel..

29. aroub

Put imaginary numbers on a side for now..

30. TuringTest

imaginary numbers are defined... now if x<0, then x^3<0 that means that (x^3)(x^3)>0 which is what we would expect given that x^6>0 since all numbers to a positive exponent are >0

31. TuringTest

another way to see it, let's assume x>0$\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}=\sqrt[10]{x^{10}}=|x|$now we do it with the negative of x$\sqrt[10]{(-x)^4} \times \sqrt[10]{(-x)^3} \times \sqrt[10]{(-x)^3}=\sqrt[10]{(-x)^4}\times\sqrt[10]{(-x)^3\cdot (-x)^3}$$=\sqrt[10]{(-x)^{10}}=\sqrt[10]{x^{10}}=|x|$so you see, after all is said and done we do not know whether the x we had at the beginning was positive or negative, hence we can only conclude that the answer is |x|

32. aroub

Okay take this as another example: $4\sqrt[4]{5a^2}\times 8\sqrt[4]{125a}\times 2\sqrt[4]{a}$ Here, a does not need an absolute value. Why? You can assume that a>0

33. akash_809

yes you would have to assume a >0

34. TuringTest

let's just look at powers of a...

35. TuringTest

if a>0$\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=|a|=a$(since we already knew a>0 now with the negative of a...

36. akash_809

yea i m waiting for that negative values of a...what would it give

37. aroub

Aoh, so you have to try the negative and the positive?

38. TuringTest

$\sqrt[4]{(-a)^2}\times\sqrt[4]{-a}\times\sqrt[4]{-a}=\sqrt[4]{a^4}=|a|\neq a$because |a|>0, but we just assumed a<0 ! a contradiction, hence the assumption makes a difference

39. TuringTest

sorry, I mean we took the negative of a...

40. TuringTest

so $|a|\neq a$

41. TuringTest

let me write that again, gotta be careful here

42. akash_809

that's the problem...the thing you did here is called fallacy of maths... http://en.wikipedia.org/wiki/Mathematical_fallacy ....search invalid roots in this article

43. TuringTest

assume a>0$\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=|a|=a$assume a<0, then$\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=|a|\neq a$since we assumed a<0

44. TuringTest

@akash_809 yes but at least one of these numbers is positive because it is raised to an even power, so I have committed no fallacy

45. akash_809

did u went through the link......the proofs given like $1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i.i=-1 ....so u hv 1=-1...and certainly \it is \not valid..$

46. TuringTest

same thing with the earlier problem...

47. akash_809

and 1=-1....surely not right...

48. TuringTest

yes, but$\sqrt{(-1)^2}\cdot\sqrt{-1}=\sqrt1\cdot\sqrt{-1}$ is not the same as$\sqrt{-1}\cdot\sqrt{-1}$

49. TuringTest

notice we have a^2 in the above, so we know *that* part is positive

50. TuringTest

this turned out to be more subtle than I bet you realized @aroub :)

51. aroub

lol, yeah true xD ALWAYS thought absolute values are so easy! And whenever someone asks me about absolute values I answer until this question came up -.-' This question is a bit confusing >.< Well, THANKS LOOAADSSSSSS :D

52. TuringTest

welcome :)